BASIC INSTRUMENTATION ELECTRICITY

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BASIC INSTRUMENTATION ELECTRICITY. 1.5V. Voltage and Current. R. R. I. Q=P/R. I=V/R. Liquid flow. Electrical current. Resistance. Every substance has resistance Conductor is substance having low resistance Isolator is substance having high resistance

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### BASIC INSTRUMENTATION ELECTRICITY

1.5V

Voltage and Current

R

R

I

Q=P/R

I=V/R

Liquid flow

Electrical current

Resistance
• Every substance has resistance
• Conductor is substance having low resistance
• Isolator is substance having high resistance
• 16 AWG wire resistance is ±12 Ω/km
• 18 AWG wire resistance is ± 20 Ω/km
• Question:
• What is the resistance of 600 m 16 AWG wire?
Voltage Drop
• When current flows across a wire the voltage will drop
• Example

I=16 mA

PT

V=24 V

length= 500 m

What is the voltage across the PT

Problem
• The allowed voltage for a Pressure Transmitter is 18V to 30 V. What is the maximum wire length if the power supply voltage in the control room is 24 V?

VAC

R

Vm

AC VOLTAGE AND CURRENT

v(t)

0

ωt

v(t) = Vmcos(ωt + )

Frequency , f = 50 Hz/ 60 Hz

T = 1/f = 1/50 = 0.02 s

ω = 2πf

 is the phase angle

v(t)

VAC

R

Vm

Im

ωt

i(t)

ωt

V and I in resistor are in phase

AC VOLTAGE and CURRENT in RESISTOR

v(t) = Vmcos ωt

i(t) = v(t)/R

= Vm (cos ωt)/R

= Imcosωt

Im = Vm/R

v(t)

ωt

i(t)

AC VOLTAGE, CURRENT and POWER in RESISTOR

v(t) = Vmcosωt

i(t) = Imcosωt

p(t) =v(t) i(t)

= VmImcos2ωt

= VmIm {1+cos(2ωt )}/2

p(t)

+

+

+

+

-

-

-

-

+ -

+ -

CAPACITOR

Unit of C is F (Farad)

1 Farad = 1 Coul/Volt = 1As/V

Real capacitor always have intrinsic capacitor and resistor with it

Im

Vm

v(t)

i(t)

ωt

0

π/2

VOLTAGE AND CURRENT IN CAPACITOR

v(t) = Vmcos(ωt +)

VAC

C

VAC

C

p(t)

POWER IN CAPACITOR

v(t)

i(t)

v(t) = Vmcos(ωt +)

+

+

p(t) = v(t) i(t)

= VmImcos(ωt + )sin(ωt + )

-

-

i(t)

coil

core

inductor

Equivalent Ckt of inductor

inductor

Unit of L is H(Henry)

1 H = 1 Vs/A

Inductor is made of coil and core.

Real inductors always have intrinsic capacitor and resistor with it

v(t)

i(t)

v(t)

Vm

Im

ωt

VOLTAGE AND CURRENT IN INDUCTOR

i(t)

for

Wehave

v(t)

p(t)

POWER IN INDUCTOR

v(t)

i(t)

v(t) =  Vmsin(ωt +)

+

+

-

p(t) = v(t) i(t)

= VmImcos(ωt + )sin(ωt + )

-

v(t)

i(t)

v(t)

i(t)

ωt

0

p(t)

+

+

-

-

Power in RL circuit

v(t)

v(t)

C

i(t)

i(t)

ωt

for

VOLTAGE AND CURRENT RCCIRCUIT

v(t)

C

i(t)

v(t)

i(t)

ωt

0

p(t)

+

+

-

-

Power in RC circuit
AC VOLTAGE, CURRENT and POWER in R, L, and C (summery )

v

v

v

i

i

i

ωt

ωt

ωt

p(t)

RESISTOR

CAPACITOR

INDUCTOR

v(t)

v(t)

i(t)

i(t)

ωt

0

ωt

0

p(t)

+

+

p(t)

+

+

-

-

-

-

Power in RC circuit

RL CIRCUIT

RC CIRCUIT

Phasors
• A phasor is a complex number that represents the magnitude and phase of a sinusoid:

VAC

R

Vm/√2

v(t)

Im /√2

i(t)

ωt

Example for V and I phasor in resistor

v(t) = Vmcos(ωt + )

i(t) = Vm/Rcos(ωt + )

VAC

C

v(t)

i(t)

Vm

Im

ωt

0

Im /√2

Vm/√2

Example for V and I phasor in capacitor

v(t) = Vmcos (ωt+)

Im/√2

Vm/√2

Example for V and I phasor in capacitor

Im/√2

Vm/√2

We can set the angle  arbitrarily. Usually we set the voltage is set to be zero phase abritrary

v(t) = Vmcos ωt

Vm/√2

Im/√2

Vm/√2

Im/√2

Example for V and I phasor in inductor

v(t)

i(t)

Here we can set the voltage to be zero phase, then the phase of current will be 

Impedance
• By definition impedance (Z) is

Z = V/I

• AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law:

V = IZ

Impedance (cont’d)
• Impedance depends on the frequency w.
• Impedance is (often) a complex number.
• Impedance is not a phasor (why?).
• Impedance allows us to use the same solution techniques for AC steady state as we use for DC steady state.
• Impedance in series/parallel can be combined as resistors

VAC

R

Vm/√2

Im /√2

Impedance of resistor

v(t) = Vmcos(ωt + )

i(t) = Vm/Rcos(ωt + )

ZR = R

VAC

C

Im /√2

Vm/√2

Impedance of capacitor

v(t) = Vmcos (ωt+)

Impedance of capacitor inductor

ωLIm/√2

v(t)

i(t)

Im/√2

ZL = jωL

Impedance

ZR = R

ZL = jωL

+

1mF

-

Impedance Example:

f = 50Hz

Find ZC

Zc = 1/jwC

w=2pf =2 × 3.14 × 50 = 314 rad/s

Zc = 1/jwC=1/(j × 314 × 106)

Zc = j3184.71

Z

ZT = Z1+ Z2

Z1

Z2

Z1

Z2

Symbol of Impedance

Impedance in series

Impedance in parallel

Impedance in series example

R = 1K2

C = 15 mF

w = 314

ZT = ?

Zc = 1/jwC=1/(j × 314 × 15 × 106)

Zc = j212.31

ZT = 1200 – j212.31

Impedance in series example

R = 1K2

ZT = ?

L = 5 mH

w = 314

ZL = jwL=j × 314 × 5 × 103

ZL = j1.57

ZT = 1200 + j1.57

R = 1K2

C = 15 mF

w = 314

Zc = j212.31

ZT = ?

L = 5 mH

w = 314

ZL = j1.57

Impedance in series example

ZT = 1200 – j212.31 + j1.57

= 1200 –j210.74

Impedance, Resistance, and Reactance

Generally impedance consist of:

The real part which is called Resistance, and

The imaginary part which is called reactance

Z = R + jX

impedance

Resistance

reactance

20kW

+

+

VC

10V  0

1mF

-

-

Example: Single Loop Circuit

w = 377 Find VC

Example (cont’d)
• How do we find VC?
• First compute impedances for resistor and capacitor:

ZR = 20kW= 20kW  0

ZC = 1/j (377 1mF) = 2.65kW  -90

20kW  0

+

+

VC

2.65kW  -90

Vi =10V  0

-

-

Impedance Example (cont’d)

Then use the voltage divider to find VC:

+

-

Impedance Example (cont’d)

20kW  0

+

VC

2.65kW  -90

Vi =10V  0

-

Complex Power

Complex power is defined as

S = VI*

The unit of complex power is Volt Ampere (VA)

S= VI* = I2Z = I2(R+jX) = I2R+jI2X

= I2Z cos  +jI2Z sin 

S = VI cos  +jVI sin  = P + jQ

Sis called apparent power and the unit is va

Pis called active power and the unit is watt and

Qis called reactive power and the unit is var

SIGNAL CONDITIONER
• Signals from sensors do not usually have suitable characteristics for display, recording, transmission, or further processing.
• They may lack the amplitude, power, level, or bandwidth required, or they may carry superimposed interference that masks the desired information.
SIGNAL CONDITIONER
• Signal conditioners, including amplifiers, adapt sensor signals to the requirements of the receiver (circuit or equipment) to which they are to be connected.
• The functions to be performed by the signal conditioner derive from the nature of both the signal and the receiver. Commonly, the receiver requires a single-ended, low-frequency (dc) voltage with low output impedance and amplitude range close to its power-supply voltage(s).
SIGNAL CONDITIONER
• Signals from sensors can be analog or digital. Digital signals come from position encoders, switches, or oscillator-based sensors connected to frequency counters.
• The amplitude for digital signals must be compatible with logic levels for the digital receiver, and their edges must be fast enough to prevent any false triggering.
• Large voltages can be attenuated by a voltage divider and slow edges can be accelerated by a Schmitt trigger.
Operational Amplifiers

The term operational amplifier or "op-amp" refers to a class of high-gain DC coupled amplifiers with two inputs and a single output. The modern integrated circuit version is typified by the famous 741 op-amp. Some of the general characteristics of the IC version are:

• High input impedance, low output impedance
• High gain, on the order of a million
• Used with split supply, usually +/- 15V
• Used with feedback, with gain determined by the feedback network.
Non-inverting Amplifier

For an ideal op-amp, the non-inverting amplifier gain is given by

Voltage Follower

The voltage follower with an ideal op amp gives simply

But this turns out to be a very useful service, because the input impedance of the op amp is very high, giving effective isolation of the output from the signal source. You draw very little power from the signal source, avoiding "loading" effects. This circuit is a useful first stage.

The voltage follower is often used for the construction of buffer for logic circuits.

Current to Voltage Amplifier

A circuit for converting small current signals (>0.01 microamps) to a more easily measured proportional voltage.

so the output voltage is given by the expression above.

Voltage-to-Current Amp

The current output through the load resistor is proportional to the input voltage